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MHT CET 2024 (2 May, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2024 (2 May, Shift 1)

Q65MHT CET 2024 · 2 MAY · SHIFT 1Easy

Moment of Inertia & Radius of Gyration

An annular ring has mass 10 kg and inner and outer radii are 5 m and 10 m respectively. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is

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Answer: (b)

  1. Annular disc about its central perpendicular axis: I=M2(R12+R22)I = \frac{M}{2}(R_1^2 + R_2^2).
  2. I=102(52+102)=5×125I = \frac{10}{2}(5^2 + 10^2) = 5 \times 125
  3. I=625 kg m2I = 625\ \text{kg m}^2
  4. So the answer is (b) 625 kg m2625\ \text{kg m}^2.
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